Mathematical Methods for Physics and Engineering : A Comprehensive Guide

31.7 HYPOTHESIS TESTING

Ten independent sample valuesxi,i=1, 2 ,..., 10 , are drawn at random from a Gaussian

distribution with standard deviationσ=1. The meanμof the distribution is known to

equal either zero or unity. The sample values are as follows:

2 .22 2.56 1.07 0.24 0.18 0.95 0. 73 − 0 .79 2.09 1. 81

Test the null hypothesisH 0 :μ=0at the10%significance level.

The restricted nature of the hypothesis space means that our null and alternative hypotheses
areH 0 :μ=0andH 1 :μ= 1 respectively. SinceH 0 andH 1 are both simple hypotheses,
the best test statistic is given by the likelihood ratio (31.108). Thus, denoting the means
byμ 0 andμ 1 , we have

t(x)=

exp

[

−^12

∑

i(xi−μ^0 )

2 ]

exp

[

−^12

∑

i(xi−μ^1 )

2 ]=

exp

[

−^12

∑

i(x

2

i−^2 μ^0 xi+μ

2

0 )

]

exp

[

−^12

∑

i(x

(^2) i− 2 μ 1 xi+μ 2
1 )

]

=exp

[

(μ 0 −μ 1 )

∑

ixi−

1

2 N(μ

2

0 −μ

2

1 )

]

.

Inserting the valuesμ 0 =0andμ 1 = 1, yieldst=exp(−N ̄x+^12 N), where ̄xis the
sample mean. Since−lntis a monotonically decreasing function oft, however, we may
equivalently use as our test statistic

v=−

1

N

lnt+^12 = ̄x,

where we have divided by the sample sizeNand added^12 for convenience. Thus we
may take the sample mean as our test statistic. From (31.13), we know that the sampling
distribution of the sample mean under our null hypothesisH 0 is the Gaussian distribution
N(μ 0 ,σ^2 /N), whereμ 0 =0,σ^2 =1andN= 10. Thus ̄x∼N(0, 0 .1).
Sincex ̄is a monotonically decreasing function oft, our best rejection region for a given
significanceαis ̄x> ̄xcrit,wherex ̄critdepends onα. Thus, in our case, ̄xcritis given by

α=1−Φ

(

x ̄crit−μ 0

σ

)

=1−Φ(10x ̄crit),

where Φ(z) is the cumulative distribution function for the standard Gaussian. For a 10%
significance level we haveα=0.1 and, from table 30.3 in subsection 30.9.1, we find
̄xcrit=0.128. Thus the rejection region onx ̄is

̄x> 0. 128.

From the sample, we deduce that ̄x=1.11, and so we can clearly reject the null hypothesis
H 0 :μ= 0 at the 10% significance level. It can, in fact, be rejected at a much higher
significance level. As revealed on p.1239, the data was generated usingμ=1.

31.7.4 The generalised likelihood-ratio test

If the null hypothesisH 0 or the alternative hypothesisH 1 is composite (or both

are composite) then the corresponding distributionsP(x|H 0 )andP(x|H 1 )are

not uniquely determined, in general, and so we cannot use the Neyman–Pearson

lemma to obtain the ‘best’ test statistict. Nevertheless, in many cases, there still

exists a general procedure for constructing a test statistictwhich has useful